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An EM wave from air enters a medium. The electric fields are E_1=E_01 x cos⁡[2π v( z -t)] in air and E_2=E_02 x cos⁡[k(2z-ct)]in medium, where the…

JEE Main 2018 — Physics Electromagnetism

2018mcqeasy

An EM wave from air enters a medium. The electric fields are E⃗1=E01x^cos⁡[2πv(zc−t)]{\vec{E}}_{1}={E}_{01}\hat{x}\mathrm{cos}⁡[2\pi v(\frac{z}{c}-t)]E1​=E01​x^cos⁡[2πv(cz​−t)] in air and E⃗2=E02x^cos⁡[k(2z−ct)]{\vec{E}}_{2}={E}_{02}\hat{x}\mathrm{cos}⁡[k(2z-ct)]E2​=E02​x^cos⁡[k(2z−ct)]in medium, where the wave number k and frequency vvv refer to their values in the air. The medium is non-magnetic. If ϵr1{\epsilon }_{{r}_{1}}ϵr1​​ and ϵr2{\epsilon }_{{r}_{2}}ϵr2​​ refer to relative permittivities of air and medium respectively, which of the following options, is correct?

Official previous-year question

Held on 8 Apr 2018 · Verified 6 Jul 2026.

Options

  1. A

    ϵr1ϵr2=12\frac{{\epsilon }_{{r}_{1}}}{{\epsilon }_{{r}_{2} }}=\frac{1}{2}ϵr2​​ϵr1​​​=21​

  2. B

    ϵr1ϵr2=4\frac{{\epsilon }_{{r}_{1}}}{{\epsilon }_{{r}_{2} }}=4ϵr2​​ϵr1​​​=4

  3. C

    ϵr1ϵr2=2\frac{{\epsilon }_{{r}_{1}}}{{\epsilon }_{{r}_{2} }}=2ϵr2​​ϵr1​​​=2

  4. D

    ϵr1ϵr2=14\frac{{\epsilon }_{{r}_{1}}}{{\epsilon }_{{r}_{2} }}=\frac{1}{4}ϵr2​​ϵr1​​​=41​

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